Given the probability distribution function of the lifetime of a battery $f(x) = \frac{2}{x^2}$ for $x > 2$ and $0$ for other $x$, out of 10 randomly selected batteries, what is the probability that at least 4 will last longer than 5 years?
So I integrated the function from 5 to infinity and got $\frac{2}{5}$ as the probability that a battery will last longer than 5 years. So I get as the probability that out of the 10 batteries, exactly 4 of the batteries will last longer than 5 years as: $(\frac{2}{5})^4 (\frac{3}{5})^6 \frac{10•9•8•7}{4•3•2•1} = 0.251...$. Now I have to do the case where 5 batteries will last longer than 5 years, 6 batteries, etc. up to 10 since the question asks for "at least." However, the total probabilities added up go over 1.0 so that can't be it. What should I do?
Your result should be $\sum_{k=4}^{10} \binom{10}{k} 0.4^k0.6^{10-k} = \frac{6032416}{9765625} \simeq 0.6177$, according to
https://www.wolframalpha.com/input/?i=sum+k%3D4+to+10+nchoosek%2810%2Ck%29%280.4%29%5Ek%280.6%29%5E%2810-k%29.
Or, if you think it is easier, it is also $1-\sum_{k=0}^{3} \binom{10}{k} 0.4^k0.6^{10-k} \simeq 0.6177$. Both will require a calculator, so I think they are of the same "difficulty."